Let $R$ be the region enclosed by the line $x=1$, the line $y=2$, the line $y=4$, and the curve $y=(x-1)^2$. $y$ $x$ ${y=(x-1)^2}$ $ 1$ $ R$ $ 2$ $ 4$ A solid is generated by rotating $R$ about the line $x=1$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=(x-1)^2}$ $ 1$ Notice the slices are horizontal, because we are rotating $R$ about a vertical axis. Each slice is a cylinder. Let the thickness of each slice be $dy$ and let the radius of the base, as a function of $y$, be $r(y)$. Then, the volume of each slice is $\pi [r(y)]^2\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(y)]^2\,dy$ This is called the disc method. What we now need is to figure out the expression of $r(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=(x-1)^2}$ $ 1$ $ 2$ $ 4$ $r$ The radius is equal to the distance between the curve ${y=(x-1)^2}$ and the line ${x=1}$. To find it, we need to solve $y=(x-1)^2$ for $x$ : ${x=\sqrt y+1}$ So, for any $y$ -value, this is the equation for $r(y)$ : $\begin{aligned} r(y)}&=({\sqrt y+1})-( 1) \\\\ &=\sqrt y} \end{aligned}$ Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(y)}]^2 \\\\ &=\pi\left(\sqrt{y}}\right)^2 \\\\ &=\pi y \end{aligned}$ The bottom endpoint of $R$ is at $y=2$ and the top endpoint is at $y=4$. So the interval of integration is $[2,4]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_2^4 \left(\pi y\right)dy \\\\ &=\pi \int_2^4y\, dy \end{aligned}$ Let's evaluate the integral. $\pi \int_2^4y\, dy=6\pi$ In conclusion, the volume of the solid is $6\pi$.